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Consider the two triangles. Triangles A B C and H G I are shown. Angles A C B and H I G are right angles. The length of side A C is 15 and the length of side C B is 20. The length of side H I is 12 and the length of I G is 9. To prove that the triangles are similar by the SAS similarity theorem, it needs to be shown that

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Answer:

Both are similar by SAS similarity.

This SAS similarity is equivalent to the congruence.

Step-by-step explanation:

Step 1:

To prove that ACB and HIG as similar triangles.

We have to look upon the corresponding sides.

SAS= Side angle sides , there the angle must be in between two sides.

[tex]\angle[/tex] ACB = [tex]\angle[/tex] HIG

Lets work on the corresponding sides.

IG/AC = IH/AC

[tex]\frac{9}{15}[/tex] = [tex]\frac{12}{20}[/tex]

Reducing each to lowest form, we divide numerator and denominator by 3 for the 1st fraction and by 4 for the 2nd fraction.

We have

[tex]\frac{3}{5}[/tex] = [tex]\frac{3}{5}[/tex]

Both sides are equal.

So its proved that both are similar with SAS similarity theorem.

Ver imagen jitushashi143

Both triangles (triangle ABC and triangle HIG) are similar by Side-Angle-Side (SAS) theorem.

Step-by-step explanation:

Given information:

Two triangle ABC and HGI are given

The length of the side AC is 15

The length of the side HI is 12

And the length of side IG is 9

Now, To prove that both triangles are similar

We have to look upon the corresponding sides

Side-Angle-Side (SAS), the angle must be in between two sides .

[tex]\angle[/tex]ACB = [tex]\angle[/tex]HIG

Now, take the corresponding sides.

IG/AC=IH/CB

[tex]9/15=12/20[/tex]

Reducing each to lowest form we get:

[tex]3/5=3/5[/tex].

Both sides are equal .

Hence it is proved that both triangles are similar by Side-Angle-Side (SAS) theorem.

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